Question: If the system of equations

\begin{align*}
2x-y&=a,\\
3y-6x &=b.
\end{align*}has a solution, find $\frac{a}{b},$ assuming $b \neq 0.$
Answer: If we multiply the first equation by $-3$, we obtain

$$3y-6x=-3a.$$Since we also know that $3y-6x=b$, we have

$$-3a=b\Rightarrow\frac{a}{b}=\boxed{-\frac{1}{3}}.$$